MATH 433
Applied Algebra
Lecture 10:
Order of a congruence class.
Fermat’s Little Theorem.
Powers of a congruence class
Let [a] ∈ Zn be a congruence class modulo n. The
powers [a]k , k = 1, 2, . . . are defined inductively:
[a]1 = [a] and [a]k = [a]k−1[a] for k > 1.
Theorem 1 [a]k+m = [a]k [a]m and [a]km = ([a]k )m
for all k, m ≥ 1.
In the case when [a] is invertible, we also let
[a]0 = [1] and [a]−k = ([a]−1)k for each k ≥ 1.
Theorem 2 If [a] is invertible, then
[a]k+m = [a]k [a]m and [a]km = ([a]k )m for all
k, m ∈ Z.
Finite multiplicative order
A congruence class [a]n is said to have finite (multiplicative)
order if [a]kn = [1]n for some positive integer k. The smallest
k with this property is called the order of [a]n . We also say
that k is the order of a modulo n.
Theorem A congruence class [a]n has finite order if and only
if it is invertible (i.e., a is coprime with n).
Proof: If [a]n has finite order k, then [1]n = [a]kn = [a]n [a]nk−1 ,
k−1
which implies that [a]−1
n = [a]n .
Conversely, suppose that [a]n is invertible. Since the set Zn is
finite, the sequence [a]n , [a]2n , [a]3n , . . . contains repetitions.
Hence for some integers 0 < r < s we will have
s
−r
[a]rn = [a]sn =⇒ [a]rn [a]−r
=⇒ [1]n = [a]s−r
n = [a]n [a]n
n .
Proposition 1 Let k be the order of an integer a modulo n.
Then as ≡ 1 mod n if and only if s is a multiple of k.
Proof: If s = k`, where ` ∈ Z, then
[as ]n = [a]sn = ([a]kn )` = [1]`n = [1]n .
Conversely, let [a]sn = [1]n . We have s = kq + r , where q is
the quotient and r is the remainder of s by k. Then
[a]r = [a]s−kq = [a]s ([a]k )−q = [1].
Since 0 ≤ r < k, it follows that r = 0.
Proposition 2 Let k be the order of an integer a modulo n.
Then as ≡ at mod n if and only if s ≡ t mod k.
Proof: If s ≡ t mod k, then s − t = `k, ` ∈ Z. It follows
that [as ] = [a]s = [a]t+`k = [a]t ([a]k )` = [a]t [1]` = [a]t = [at ].
Conversely, if [as ] = [at ], then
[a]s−t = [a]s [a]−t = [a]s ([a]t )−1 = [as ][at ]−1 = [1].
By Proposition 1, s − t is a multiple of k.
Examples. • G7 = {[1], [2], [3], [4], [5], [6]}.
[1]1 = [1],
[2]2 = [4], [2]3 = [8] = [1],
[3]2 = [9] = [2], [3]3 = [2][3] = [6], [3]4 = [2]2 = [4],
[3]5 = [4][3] = [5], [3]6 = [3][5] = [1].
[4]2 = [16] = [2], [4]3 = [4][2] = [1].
[5]2 = [25] = [4], [5]3 = [4][5] = [−1], [5]4 = [−1][5] = [2],
[5]5 = [2][5] = [3], [5]6 = [3][5] = [1].
[6]2 = [−1]2 = [1].
Thus [1] has order 1, [6] has order 2, [2] and [4] have order 3,
and [3] and [5] have order 6.
• G12 = {[1], [5], [7], [11]}.
[1]1 = [1], [5]2 = [25] = [1], [7]2 = [−5]2 = [25] = [1],
[11]2 = [−1]2 = [1].
Thus [1] has order 1 while [5], [7], and [11] have order 2.
Fermat’s Little Theorem Let p be a prime number. Then
ap−1 ≡ 1 mod p for every integer a not divisible by p.
Proof: Consider two lists of congruence classes modulo p:
[1], [2], . . . , [p − 1] and [a][1], [a][2], . . . , [a][p − 1].
The first one is the list of all elements of Gp . Since a is not a
multiple of p, it’s class [a] is in Gp as well. Hence the second
list also consists of elements from Gp . Also, all elements in
the second list are distinct as
[a][n] = [a][m] =⇒ [a]−1 [a][n] = [a]−1 [a][m] =⇒ [n] = [m].
It follows that the second list consists of the same elements as
the first (arranged in a different way). Therefore
[a][1] · [a][2] · · · [a][p − 1] = [1] · [2] · · · [p − 1].
Hence [a]p−1 X = X , where X = [1] · [2] · · · [p − 1].
Note that X ∈ Gp since Gp is closed under multiplication.
That is, X is invertible. Then [a]p−1 XX −1 = XX −1
=⇒ [a]p−1 [1] = [1] =⇒ [ap−1 ] = [1].
Corollary 1 Let p be a prime number. Then ap ≡ a mod p
for every integer a (that is, ap − a is a multiple of p).
Corollary 2 Let a be an integer not divisible by a prime
number p. Then the order of a modulo p is a divisor of p − 1.
Proof: By Fermat’s Little Theorem, ap−1 ≡ 1 mod p.
According to a previously proved proposition, the order of a
modulo p divides any positive integer s such that
as ≡ 1 mod p.
Problem. Find the remainder of 1250 under division by 17.
Since 17 is prime and 12 is not a multiple of 17, we have
50
50
[12]16
= [12]3·16+2 =
17 = [1]17 . Then [12 ] = [12]
= ([12]16 )3 · [12]2 = [12]2 = [−5]2 = [25] = [8]. Hence the
remainder is 8.